Math 1A Section 3231 Fall 2016
Instructor: John Kwon
Extra Credit Challenge #2 - Solution
1A. Let f be a function defined by
f (x) = lim
t→x
sec t − sec x
.
t−x
Find the exact value of f 0 (π/4) .
Solution. Let g(t) = sec t. Applying the definition of the derivative at t = x, we have
g 0 (x) = lim
t→x
Therefore
f (x) = g 0 (x) =
sec t − sec x
.
t−x
d
sec x = sec x tan x ,
dx
and, by the Product Rule,
f 0 (x) = sec x · sec2 x + sec x tan x · tan x = sec x(sec2 x + tan2 x) .
Finally, from the trigonometric identity 1 + tan2 x = sec2 x, we obtain tan2 x = sec2 x − 1 . Using
this identity, we eliminate tan x in f 0 (x) and write everything in terms of sec x :
f 0 (x) = sec x(sec2 x + sec2 x − 1) = sec x(2 sec2 x − 1) .
Therefore
π π 2
i
√ h
√
√
f (π/4) = sec
2 · sec
− 1 = 2 2 · ( 2)2 − 1 = 3 2 .
4
4
0
1B. Show that
| sin x − cos x| ≤
√
2.
(Hint: Maximize and minimize the function f (x) = sin x − cos x .)
Solution. First of all, note that both sin x and cos x are continuous at every real number,
so f (x) = sin x − cos x is continuous on all of R. Moreover, since sin(x + 2π) = sin x and
cos(x + 2π) = cos x for all x ∈ R, f is a periodic function with the period no greater than 2π.
Therefore, f will have its absolute maximum and minimum on the closed interval [0, 2π] (as guaranteed by the Extreme Value Theorem), and this maximum and minimum will be the maximum
and minimum of f on all of the real numbers.
The critical points of f on [0, 2π] can be found be solving the equation
f 0 (x) = cos x + sin x = 0
on [0, 2π], from which we obtain
sin x = − cos x .
If cos x 6= 0, then we can divide both sides by cos x and the equation becomes
sin x
= −1
cos x
=⇒
tan x = −1
=⇒
x=
3π
7π
or x =
.
4
4
These are the critical points of f (x) = sin x − cos x on the interval [0, 2π]. Comparing the values
of f (x) at the critical points and the endpoints of the interval [0, 2π], we have
x
f (x) = sin x − cos x
0
f (0) = sin 0 − cos 0 = −1
3π/4
7π/4
2π
If cos x = 0, then we have x =
f (3π/4) = sin(3π/4) − cos(3π/4) =
√
2
√
f (7π/4) = sin(7π/4) − cos(7π/4) = − 2
f (2π) = sin 2π − cos 2π = −1
π
3π
or x =
, but we see that
2
2
f (π/2) = sin(π/2) − cos(π/2) = 1 <
√
2
√
f (3π/2) = sin(3π/2) − cos(3π/2) = −1 > − 2
√
√
so the absolute minimum and absolute maximum values of f (x) on [0, 2π] are − 2 and 2,
respectively. Since f (x) is a periodic function with period no greater than 2π, this means that
√
√
− 2 ≤ f (x) ≤ 2
which implies
|f (x)| ≤
√
2
for all x ∈ R, which is the desired result.
2A. The figure below shows a circle with radius 1 and the parabola y = x2 so that they share
the common tangent lines in two points. Find the center of the circle.
Solution. Consider the close-up view in the first quadrant in the figure above. Let P (x0 , y0 )
be the point where the circle and the parabola tangentially meet, and consider two more points,
A(0, y0 ) and B(0, b), as shown. Our goal is to find (0, b).
The key observation is that the circle x2 + (y − b)2 = 1 and the parabola y = x2 share the
common tangent line at P , hence having the same slope of the tangent line. Let’s find the slope of
the tangent line of the circle at the point P (x0 , y0 ). Differentiating the equation x2 + (y − b)2 = 1
implicitly with respect to x gives
2x + 2(y − b)
Substituting x = x0 and y = y0 yields
at the point P .
dy
=0
dx
=⇒
dy
x
=
dx
b−y
x0
, which is the slope of the tangent line of the circle
b − y0
Likewise, the slope of the tangent line of the parabola is given by taking the derivative of y = x2 :
dy
= 2x
dx
so the slope of the tangent line of the parabola at P (x0 , y0 ) is 2x0 . Equating the two slopes, we
now obtain
x0
= 2x0 .
b − y0
Note that P (x0 , y0 ) is clearly in the first quadrant, so x0 6= 0, which allows the division of both
sides of this equation by x0 , which yields:
1
= 2
b − y0
(Continued on the next page)
which implies that
b − y0 =
1
.
2
1
Note that AB = b − y0 = . Since 4P AB is a right triangle, by Pythagorean Theorem we see
2
that
√
p
p
3
2
2
2
2
AP =
P B − AB =
1 − (1/2) =
.
2
√
3
. Since (x0 , y0 ) is on the parabola y = x2 , we have y0 = x0 2 .
Since AP = x0 , we see that x0 =
2
Therefore
√ !2
3
3
y0 =
= .
2
4
Then, from b − y0 =
1
, we calculate
2
3 1
5
1
=
+
= .
2
4 2
4
5
.
Therefore the center of the circle is 0,
4
b = y0 +
2B. Let y = mx + b be a line, with b > 0, so that it intersects the parabola y = x2 in points
A and B, as shown below. Find the point P on the arc AOB of the parabola that maximizes
the area of the triangle P AB.
Solution. Let the x- and y-coordinates of the point P be x and y, respectively. We want to
maximize the area of 4P AB, so first we need to express this area as a function of one variable.
Let x1 and x2 denote the x-coordinates of the points A and B, respectively. Since A and B are
the points of intersection between the line y = mx + b and the parabola y = x2 , we see that x1
and x2 are constants that depend on the values of the constants m and b. For now, noticing that
x1 and x2 are constants is sufficient; we will be able to find them in terms of m and b later if we
need to. Finally, note that it is clear from the picture that x1 < x2 and x1 ≤ x ≤ x2 .
Draw the vertical line through P , as shown in the figure above, and let C denote the intersection
between this vertical line and the line y = mx + b. Then we see that
Area(4P AB) = Area(4CP A) + Area(4CP B) .
Note that, if we view CP as the base of 4CP A, then its height is (x − x1 ). It is clear that
CP = mx + b − x2 . Therefore
Area(4CP A) =
1
(mx + b − x2 )(x − x1 ) .
2
Likewise, if we view CP as the base of 4CP B, then its height is (x2 − x). Hence
Area(4CP B) =
1
(mx + b − x2 )(x2 − x) .
2
Adding the two areas, we obtain
1
1
(mx + b − x2 )(x − x1 ) + (mx + b − x2 )(x2 − x)
2
2
1
x
−
x
2
1
= (mx + b − x2 )(x − x1 + x2 − x) =
(mx + b − x2 ) .
2
2
Area(4P AB) = Area(4CP A) + Area(4CP B) =
(Continued on the next page)
Therefore, the function
x2 − x1
(mx + b − x2 )
2
represents the area of 4P AB as a function of x, where m, b, x1 , and x2 are all constants. Then
A(x) =
A0 (x) =
x2 − x1
(m − 2x) .
2
Setting this equal to 0 and solving for x yields the critical point of A(x) :
x2 − x1
(m − 2x) = 0
2
=⇒
x=
m
.
2
Since
x2 − x1
· (−2) = x1 − x2 < 0 ,
2
m
by the Second Derivative Test we see that the critical point x =
yields a local maximum.
2
Finally, since this local maximum is the only local extremum for A(x), by Theorem 4.5 we see
m
m
that A(x) attains its absolute maximum at x =
as well. Hence
is the x-coordinate of the
2
2
point P that maximizes the area of 4P AB.
A00 (x) =
Since the point P lies on y = x2 , the y-coordinate of P is
that maximizes the area of 4P AB is
m m2
,
2
4
m 2
2
=
m2
. Therefore the point P
4
.
(Note that we didn’t need to find x1 and x2 in terms of m and b, after all!)